Problem 203 Answer

Question

The game of one-card poker has a three-card deck, an ace, deuce, and trey. The ace is lowest and the trey is highest. Each of two players ante $1 into the pot. Then, each player gets one card. The order of betting is predetermined, with player 1 to act first. Player 1 may either bet $1 or check. If player 1 bets, player 2 may either call or fold. If player 1 checks, player 2 may either bet $1 or check. If player 1 checks, and player 2 bets, then player 1 may either call or fold. If both players check, or one bets and one calls, then the higher card wins the pot. Assuming both players are perfect logicians, what is the optimal strategy for each player?

Solution

The obvious plays are neither player should call with the ace, always call with the trey, and never bet with a deuce. By raising with a deuce, you will win $1 against an ace, because your opponent will fold, and lose $2 against a trey, because he will raise. A superior strategy with a deuce is to always check, and then to always call if the other player bets. By following the check and call strategy, you'll win $2 instead of $1 against an ace, and stay equal against a trey at a $2 loss. If your opponent checks with the ace, you'll still win $1, and be no worse off. You can only gain by checking and calling with a deuce.

There are five other, more difficult, decision points to be faced by both players, as follows:

p₁

Probability player 1 bets with an ace (bluffing).

p₂

Probability player 1 calls with a deuce after player 2 bets (hoping player 2 is bluffing).

p₃

Probability player 1 bets with a trey (hoping player 2 will call).

q₁

Probability player 2 bets with an ace, after player 1 checks (bluffing).

q₂

Probability player 2 calls with a deuce, after player 1 bets (hoping player 1 is bluffing).

The key to game theory problems like this is to put your opponent to an indifference point at non-obvious decisions. If the odds favor one way or the other, the person making the decision will always go with the odds, and gain an advantage. To do that, let's express player 1's expected value.

EV1 =
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 checks)×pr(player 2 checks)×-1 +
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 folds)×+1 +
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 calls)×-2 +
pr(player 1 has A)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×-1 +
pr(player 1 has A)×pr(player 2 has 3)×pr(player 1 bets)×pr(player 2 calls)×-2 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×-1 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×+2 +
pr(player 1 has 2)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×-1 +
pr(player 1 has 2)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×-2 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×+2 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 bets)×pr(player 2 folds)×+1
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 folds)×+1
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 calls)×+2

EV1 =
(1/3)×(1/2)×(1-p₁)×1×-1 +
(1/3)×(1/2)×p₁×(1-q₂)×1 +
(1/3)×(1/2)×p₁×q₂×-2 +
(1/3)×(1/2)×(1-p₁)×1×1×-1 +
(1/3)×(1/2)×p₁×1×-2 +
(1/3)×(1/2)×1×(1-q₁)×1 +
(1/3)×(1/2)×1×q₁×(1-p₂)×-1 +
(1/3)×(1/2)×1×q₁×p₂×2 +
(1/3)×(1/2)×1×1×(1-p₂)×-1 +
(1/3)×(1/2)×1×1×p₂×-2 +
(1/3)×(1/2)×(1-p₃)×(1-q₁)×1 +
(1/3)×(1/2)×(1-p₃)×q₁×1×2 +
(1/3)×(1/2)×p₃×1×1 +
(1/3)×(1/2)×(1-p₃)×1×1 +
(1/3)×(1/2)×p₃×(1-q₂)×1 +
(1/3)×(1/2)×p₃×q₂×2

EV1 = (1/6)×( -1+p₁ +p₁-p₁q₂ -2p₁q₂ -1+p₁ -2p₁ +1-q₁ -q₁+q₁p₂ +2q₁p₂ -1+p₂ -2p₂ 1-q₁-p₃+p₃q₁ +2q₁-2p₃q₁ +p₃ +1-p₃ +p₃-p₃q₂ +2p₃q₂)

= (1/6)×(p₁-q₁-p₂-3p₁q₂+3q₁p₂-p₃q₁+p₃q₂) Equation 1
= (1/6)×[q₁×(-1+3p₂-p₃) + q₂×(-3p₁+p₃) + p₁-p₂]

What values for q₁ and q₂ will make EV1 equal to a constant, causing player 1 to be indifferent to all values of p₁, p₂, and p₃?

The answer is q₁=1/3, and q₂=1/3, so...

EV1 = (1/6)×[(1/3)×(-1+3p₂-p₃) + (1/3)×(-3p₁+p₃) + p₁-p₂]
= (1/6)×[-1/3 +p₂ -p₃/3 - p₁ +p₃/3 +p₁ -p₂]
= (1/6)×(-1/3)
= -1/18

So, player 2 should bluff with an ace after a check, and call with a deuce, each with probability 1/3. If player 2 does that, it won't make any difference what player 1 does at his three difficult decision points, his overall expected value will be -1/18.

Next, what should player's 1 optimal strategy be, so that player 2 can't exploit a bad strategy? Let's go back to equation 1. The expected value of player 2 will be the opposite of the expected value of player 1. So...

EV2 = (-1/6)×(p₁-q₁-p₂-3p₁q₂+3q₁p₂-p₃q₁+p₃q₂) =
EV2 = (-1/6)×[p₁×(1-3q₂)+p₂×(3q₁-1)+p₃×(q₂-q₁)-q₁]

Next, determine values for p₁, p₂, and p₃ so that the q₁ and q₂ terms cancel out, leaving a constant. This will result in a strategy where player 2 is indifferent at the difficult decision points.

If 3p₂-p₃=1, then the q₁ terms will cancel out. If 3p₁=p₃, then then q₂ terms will cancel out.

So, two equations must be satisfied. Examples of values that work are p₁=1/5, p₂=8/15, p₃=3/5.

Acknowledgement: The idea of this problem came from the article Game Theory and Poker, by Jason Swanson (April 3, 2005). The paper works out the same solution. I kept the variables names the same for my own benefit, and yours.